Low-dimensional representations of the three component loop braid

LOW-DIMENSIONAL REPRESENTATIONS OF THE THREE COMPONENT LOOP BRAID GROUP

arXiv:1508.00005v1 [math.RT] 31 Jul 2015

PAUL BRUILLARD, SEUNG-MOON HONG, JULIA YAEL PLAVNIK, ERIC C. ROWELL, LIANG CHANG, AND MICHAEL YUAN SUN Abstract. Motivated by physical and topological applications, we study representations of the group LB 3 of motions of 3 unlinked oriented circles in R3 . Our point of view is to regard the three strand braid group B3 as a subgroup of LB 3 and study the problem of extending B3 representations. We introduce the notion of a standard extension and characterize B3 representations admiting such an extension. In particular we show, using a classification result of Tuba and Wenzl, that every irreducible B3 representation of dimension at most 5 has a (standard) extension. We show that this result is sharp by exhibiting an irreducible 6-dimensional B3 representation that has no extensions (standard or otherwise). We obtain complete classifications of (1) irreducible 2-dimensional LB 3 representations (2) extensions of irreducible 3-dimensional B3 representations and (3) irreducible LB 3 representations whose restriction to B3 has abelian image.

1. Introduction and Motivation Over the last two decades, topological states of matter in 2 spatial dimensions and their potential computational applications have motivated the study of motions of sytems of pointlike excitations on 2-dimensional surfaces. The mathematical model for such systems are (2 + 1)-topological quantum field theories (TQFTs). The motions of points in the disk lead to representations of the braid group Bn which play a central role in the topological model for quantum computation [8, 9]. Recently the possibility of 3-dimensional topological states of matter (see, e.g. [25, 26]) modeled by (3 + 1)-TQFTs [24] lead to new possible avenues for quantum computation. Although motions of point-like excitations in 3 spatial dimensions are mathematically trivial, the symmetries of loop-like excitations can be quite complicated. The simplest mathematical manifestation of this idea is the group LB n of motions of an oriented n-component unlink Cn [5, 11, 18]. In this article, we begin a systematic study of the low-dimensional representations of the loop braid group LB3 . The generators and relations for LB3 are given in [6]. We take it as a definition here: Definition 1.1. The three component loop braid group LB3 is the abstract group generated by σ1 , σ2 , s1 , s2 satisfying the following relations: PNNL Information Release: PNNL-SA-111814. This paper began while the authors were participating in an AMS Mathematics Research Community workshop in Snowbird, Utah. The continued support of the AMS is gratefully acknowledged. J.Y.P was partially supported by CONICET, ANPCyT and Secyt (UNC). E.C.R. was partially supported by NSF DMS-1108725. M.Y.S. acknowledges the support of SFB878 and GIF. The research described in this paper was, in part, conducted under the Laboratory Directed Research and Development Program at PNNL, a multi-program national laboratory operated by Battelle for the U.S. Department of Energy. 1

(B1) (S1) (S2) (L1) (L2)

σ1 σ2 σ1 = σ2 σ1 σ2 s1 s2 s1 = s2 s1 s2 s21 = s22 = 1 s1 s2 σ1 = σ2 s1 s2 σ1 σ2 s1 = s2 σ1 σ2

The group satisfying only (B1), (S1) and (S2) is the free product B3 ∗ S3 of Artin’s braid group B3 and the symmetric group S3 . The group satisfying (B1)-(L1) is known as the virtual braid group on three strands VB 3 [14]. Some authors replace (L2) by (L2′ ) σ2 σ1 s2 = s1 σ2 σ1

which leads to an isomorphic group. The group satisfying (B1)-(L2) and (L2′ ) is called the symmetric loop braid group SLB 3 [12]. The 3-component loop braid group LB3 is geometrically understood as motions of 3 oriented circles in R3 . The generator σi is interpreted as passing the ith circle under and through the i + 1st circle ending with the two circles’ positions interchanged. The generator si corresponds to simply interchanging circles i and i + 1. We can represent these generators diagrammatically as follows, with the time variable to be read from bottom to top:

σ1 =

σ2 =

1

2

3

1

2

3

1

2

3

s2 =

s1 =

1

2

3

A few authors have approached the representations of LB3 from the point of view of extending representations of B3 (see [1, 2, 3, 12, 23]). Among these, the extending of specific families of representations of B3 , such as the Burau and Lawrence-Krammer-Bigelow representations, as well as certain representations obtained from solutions to the Yang-Baxter equation have been considered. We will pay special attention to extensions of the Tuba-Wenzl [22] representations, which exhaust all of the irreducible representations of B3 in five dimensions or less. In doing so we obtain many insights into finite-dimensional representations of LB3 , especially in lower dimensions. Results and Methodology. Our basic strategy is the following: (1) Let (ρ, V ) be a representation of B3 , with ρ(σ1 ) = A and ρ(σ2 ) = B. 2

(2) Find S, S1 , S2 ∈ End(V ) so that ρ(s1 ) = S1 , ρ(s2 ) = S1 S and S = S1 S2 extends ρ to a representation of VB 3 (see for example Theorem 2.4 and Proposition 2.11). (3) Determine which pairs (S1 , S2 ) from the previous step factor over LB3 . Our original goal was to carry out this procedure with an irreducible representation in step (1). Indeed, the classification results of [22] are very explicit: they show that any irreducible B3 representation of dimension 5 or less is equivalent to a matrix representation with A and B in ordered triangular form: A (respectively, B) is upper (respectively, lower) triangular and Bi,i = Ad−i+1,d−i+1 . This is accomplished as part of the following theorem: Theorem. Let ρ : B3 → GLd (C) be an irreducible representation of B3 . • If d ≤ 5, then there exists an extension of ρ to LB3 . • If d = 6, there is an irreducible representation of B3 with no possible extensions.

We show the above theorem using the notion of a standard extension, which are those for which ρ(s1 s2 ) = S = kAB for some k ∈ C. The main advantage of these extensions is that they trivialize step (3) in our methodology (see Lemma 2.2). We have gone somewhat beyond our original goal, which we now summarize. Summary of results. (1) In Section 2 we characterize when standard extensions exist (Theorem 2.4). We show all irreducible representations of LB3 when ρ(σ1 ) = A = B = ρ(σ2 ) arise as standard extensions (Theorem 2.8). We also show that extensions which factor through SLB 3 are rare (Theorem 2.16). An infinite-dimensional is given. (2) We have determined all two-dimensional representations of LB3 in Section 3. In particular, we show in Theorem 3.1 that every irreducible two-dimensional representation of LB3 is a standard extension of some B3 representation and every two-dimensional B3 representation admits a standard extension. (3) In Section 4, we classify extensions of irreducible three dimensional B3 representations and show that, generically, irreducible three dimensional representations of LB3 are standard extensions. (4) In Section 5, we show all four and five-dimensional Tuba-Wenzl representations, including the reducible ones, admit standard extensions. (5) In Section 6, we give an example of an irreducible B3 representation that has no possible extensions to LB3 , and give some evidence for a conjecture that any B3 representation in the Tuba-Wenzl ordered-triangular form must have an extension. 2. General results We record here our main results on extending B3 representations to LB3 , which are not dimension specific. We first introduce the notion of a standard extension. Throughout this section, and for the remainder of the text, V will be a finite dimensional vector space over C unless otherwise stated. 2.1. The standard extension. Definition 2.1. A standard extension of a representation ρ : B3 → GL(V ) to LB 3 is one for which ρ(s1 s2 ) = kρ(σ1 σ2 ) for some k ∈ C. The following lemma shows that if A, B, S1 and S2 satisfy relations (B1) and (S1) and S1 S2 is proportional to AB, then (L1) and (L2) are also satisfied. 3

Lemma 2.2. Let A, B, S1 , S2 ∈ End(V ) satisfy ABA = BAB and S1 S2 S1 = S2 S1 S2 . If S1 S2 = kAB for some k ∈ C× then ABS1 = S2 AB and S1 S2 A = BS1 S2 . Proof. ABS1 = k −1 S1 S2 S1 = k −1 S2 S1 S2 = S2 AB, S1 S2 A = kABA = kBAB = BS1 S2 .  Next we describe extensions of representations of the alternating group A3 to S3 . We use cycle notation for elements of S3 , and denote by ω a primitive 3rd root of unity. Lemma 2.3. Let γ : A3 → GL(V ) be a representation given by γ((1 2 3)) = S. Then γ can be extended to S3 by γ(s1 ) = S1 and γ(s2 ) = S2 with γ(s1 s2 ) = S1 S2 = S if and only if Tr(S) ∈ Z. Denoting by Vλ the λ-eigenspace of S, we can take S1 to be any involution which preserves V1 and interchanges Vω and Vω−1 . Proof. If γ is a representation of S3 then Tr(s1 s2 ) ∈ Z, since every character of S3 is integral. Conversely, if Tr(S) ∈ Z then the non-real eigenvalues ω ±1 of S must appear with the same multiplicity. Now take S1 to be any involution that preserves the 1-eigenspace of S and interchanges the ω and ω −1 eigenspaces, and define S2 = S1 S. It is enough to verify that S1 S2 S1 = S2 S1 S2 , whence S22 = I will follow from (S1 S2 S1 )2 = S 3 = I. We verify the equivalent condition SS1 = S1 S 2 . On V1 this is clearly true. For v ∈ Vω , we have S1 S 2 v = ω −1 S1 v and S1 v ∈ Vω−1 so SS1 v = ω −1 (S1 v). Conversely, the relation SS1 = S1 S 2 with the same argument in reverse shows that the involution S1 must preserve  V1 and interchange vectors in Vω and Vω−1 . We now characterize when a B3 representation has a standard extension. Theorem 2.4. Let ρ : B3 → GL(V ) be a finite-dimensional representation of B3 , with ρ(σ1 ) = A and ρ(σ2 ) = B. Then the following are equivalent for k and m in C (a) ρ has a standard extension to LB3 with S = kAB and Tr(S) = m, (b) (AB)3 = k −3 I and Tr(AB) = k −1 m with m ∈ Z. (c) AB is diagonalizable and Tr((AB)ℓ ) = k −ℓ m with m ∈ Z for all ℓ ≤ dim V not divisible by 3 and Tr((AB)ℓ ) = k −ℓ dim V for all ℓ ≤ dim V divisible by 3. Proof. The equivalence of (a) and (b) can be seen from Lemmas 2.2 and 2.3. The implication from (a) to (c) is obtained by taking the trace of both sides of S ℓ = (kAB)ℓ . For (c) implies (b), we appeal to the fact that the coefficients of the characteristic polynomial of a d × d matrix X is determined by the numbers Tr(X ℓ ) for 1 ≤ ℓ ≤ d (see for example [20] or [27]). In particular the characteristic polynomial of AB is identical to the characteristic polynomial of some matrix X satisfying X 3 = k −3 I and Tr(X) = k −1 m. Therefore AB and X have the same set of eigenvalues, and since AB is assumed to be diagonalizable we are done.  Remark 2.5. When Tr(kAB) 6= 0, there is at most one possible value of k for which Tr(kAB) ∈ Z, otherwise there are exactly three such values differing by a factor of ω. In any case, we can now produce all standard extensions of a given B3 representation satisfying either condition (b) or (c) of Theorem 2.4. (1) For fixed ρ(σ1 ) = A, ρ(σ2 ) = B, choose k ∈ C so that (AB)3 = k −3 I and Tr(kAB) ∈ Z. Define S = kAB. (2) Choose any M so that M −1 SM = (Iℓ , ωIt , ω 2It ). (3) Pick any G ∈ GLt (C) and 0 ≤ a ≤ ℓ and an N ∈ GLℓ (C). 4

  N(Ia , −Iℓ−a )N −1 0 0 0 0 G . (4) Set S1 =  0 G−1 0 (5) Define ρ(s1 ) = MS1 M −1 and ρ(s2 ) = ρ(s1 )S. Notice that distinct involutions on Cℓ with characteristic polynomial (x − 1)a (x + 1)ℓ−a are parameterized by N(Ia , −Iℓ−a )N −1 , where N ∈ GLℓ (C)/(GLa (C) × GLℓ−a (C)). Denote by ⌈n⌉ = min{k ∈ Z : n ≤ k} the usual ceiling function. Proposition 2.6. Let ρ : LB3 → GL(V ) be a representation with ρ(σ1 ) = A, ρ(σ2 ) = B and ρ(s1 s2 ) = S = kAB. Suppose W is invariant under A and B. Then W is LB3 invariant only if dim W ∩ Vω = dim W ∩ Vω2 . In this case, denote dim W ∩ Vω = dim W ∩ Vω2 = m and let dim W ∩ V1 = ℓ. For ρ(s1 ) = S1 and ρ(s2 ) = S2 we have: (a) (S1 |W , S2 |W ) are parametrized by GLm (C) × ∐ℓj=0 (GLℓ (C)/(GLj (C) × GLℓ−j (C))). (b) Unitary (S1 |W , S2 |W ) are parametrized by Um × ∐ℓj=0 (Uℓ /(Uj × Uℓ−j )). 2 (c) In either case, the parameter space has dimension m2 (⌈ ℓ 2−1 ⌉) if ℓ > 1, and m2 otherwise.

Proof. For (a) we apply Lemma 2.3 to W and use Remark 2.5. A small modification gives (b). Part (c) is a routine calculation.  2.2. Extensions where ρ(σ1 ) = ρ(σ2 ). We classify here the irreducible LB3 representations V in which A = ρ(σ1 ) = ρ(σ2 ) = B. In this case the mixed relation (L1) implies S := ρ(s1 s2 ) commutes with A. Thus the S-eigenspace decomposition V = V1 ⊕ Vω ⊕ Vω2 must be Astable. Moreover, setting S1 := ρ(s1 ) and S2 = ρ(s2 ) = S1 S and restricting to the subgroup hs1 , s2 i ∼ = S3 we see that V1 and Vω ⊕ Vω2 are complementary subrepresentations of V . Therefore it suffices to assume either V = V1 or V = Vω ⊕ Vω2 . We consider the first case in Theorem 2.7 and the second in Theorem 2.8. Theorem 2.7. Any irreducible LB3 representation V with ρ(s1 s2 ) = I and A = ρ(σ1 ) = ρ(σ2 ) = B and dim(V ) > 1 is 2-dimensional of the form     λ x 0 1 A= ρ(s1 ) = S1 = ρ(s2 ) = S2 = , 0 −λ 1 0 for x ∈ C and λ ∈ C× . No two are isomorphic.



Proof. Since V = V1 , S1 = S2 . The group G := ha, b : a2 = 1, [a, b2 ] = 1i is isomorphic to a split extension K ⋊ Z2 where K is the kernel of G → Z2 sending both a an b to 1 ∈ Z2 . Clearly ρ(LB3 ) = hA, S1 i is a quotient of G ∼ = K ⋊ Z2 under which the image of K is 2 2 L := hA , AS1 , S1 Ai. Note that A ∈ Z(L) and AS1 = (S1 A)−1 modulo Z(L). Hence we have 1 → Z → L → Z → 1, which must split to give L = Z × Z. Therefore our representation factors over (Z × Z) ⋊ Z2 . We show it must be at most two-dimensional and leave the rest to the reader: restrict to the abelian subgroup ρ(K) and consider a one-dimensional subrepresentation C. By Frobenius reciprocity, HomLB3 (ind C, V ) = HomK (C, res V ) 6= 0, which means there is a non-zero map from ind C to V . Since V is irreducible and ind C is two-dimensional then V must be at most two-dimensional.  5

Theorem 2.8. Let (ρ, V ) be an irreducible representation of LB 3 such that A = ρ(σ1 ) = ρ(σ2 ) = B and V1 = 0. If V is finite-dimensional, then dim V = 2n and there is a basis such that   0 In −1 2 ρ(s1 s2 ) = S = µ ωA , ρ(s2 ) = S2 = , A = diag(A1 , A2 ) In 0 √  √  µ µ   0 µ     0 µ     1 0     1 0     0 µ .     .. A1 =  1 0  or       . 0 µ     ..     1 0   √ 0 µ ∓ µ 1 0     0 µω 0 µω 1 0     1 0     0 µω     0 µω     1 0     1 0 A2 =  or ..    . .     ..     0 µω    0 µω    1 0 √ 2 1 0 µω Proof. Suppose V = Vω ⊕ Vω2 . Over C, there is some non-zero A2 -eigenspace Vω,µ ⊂ Vω . The relation A2 S = S2 A2 S2 shows S2 (Vω,µ ) = Vω2 ,µω . Both eigenspaces are preserved by A and so if V is irreducible, we can assume V = Vω,µ ⊕ Vω2 ,µω and that there are no proper A invariant subspaces of Vω,µ whose image under S2 is A-invariant. Let v ∈ Vω,µ be an eigenvector of A and consider the sequence v, S2 v, AS2 v, S2 AS2 v, . . . , (AS2 )l v, S2 (AS2 )l v, . . . . Since V is finite-dimensional, this sequence will eventually be linearly dependent. Let n be the largest integer for which the first 2n terms are linearly independent. Note any odd number of independent terms cannot be maximal because S2 is a linear bijection between those terms in Vω,µ and those in Vω2 ,µω . Therefore the 2n + 1 term (AS2 )2n v is in the span of the previous 2n terms, which in particular means the span of the first 2n terms are A invariant. Since it was already obviously S2 and S invariant, we have the span of the first 2n terms is all of V . We also have that those terms with an even number of S2 ’s appearing span Vω,µ and those with an odd number span Vω2 ,ωµ . We see with respect to this basis, the matrices have the desired form.  Example 2.9. Let V be an infinite-dimensional vector space with basis {e1 , f1 , e2 , f2 , . . . }. Then we can define a representation of LB3 with respect to some µ ∈ C× and A = B, V1 = 0: Sei = ωei , Sfi = ω 2 fi , S2 ei = fi ,

Ae1 = µe1 , Aei = ei+1 , Afi−1 = fi even i. Aei = µ2 ei−1 , Afi+1 = µ2 fi odd i. 6

2.3. Beyond the standard extension. The following computational result will be useful: Lemma 2.10. Suppose S1 , S2 ∈ GLd (C) satisfy (S1) and (S2), i.e. S1 S2 S1 = S2 S1 S2 and Si2 = I. Then the following are equivalent for matrices A, B ∈ GLd (C): (a) ABS1 = S2 AB, (b) S2 commutes with ABS −1 [or equivalently with S(AB)−1 ], (c) S1 commutes with (AB)−1 S [or equivalently with S −1 (AB)], (d) ABS = S2 ABS2 Proof. (a) implies (b): Then ABS −1 S2 = ABS2 S1 S2 = ABS1 S2 S1 = S2 ABS2 S1 = S2 ABS −1 . (b) implies (c): (AB)−1 SS1 = (AB)−1 S2 S1 S2 = S −1 S2 S(AB)−1 S = S1 (AB)−1 S. (c) implies (d): ABS = SS −1 (AB)S1 S2 = SS1 S −1 (AB)S2 = S2 ABS2 (d) implies (a): ABS1 = ABSS2 = S2 ABS2 S2 = S2 AB.  Proposition 2.11. Let ρ be a B3 representation with ρ(σ1 ) = A and ρ(σ2 ) = B. Suppose there is an extension ρ′ (s1 ) = S1′ and ρ′ (s2 ) = S2′ , with ρ′ (s1 s2 ) = S ′ to LB3 . Suppose further that a standard extension is possible for k ∈ C× . Then ρ(s1 s2 ) = S = kB 2 S ′ defines an extension of ρ to VB 3 . In particular, S = k 2 B 2 AB gives VB 3 representations. Proof. We show Tr(S) ∈ Z, SA = BS and S 3 = I and apply Lemma 2.3. First we show S satisfies SA = BS: SA = kB 2 S ′ A = kB 2 BS ′ = BkB 2 S ′ = BS. Now we check the trace of S using Lemma 2.10(d): Tr(S) = kTr(B 2 S ′ ) = kTr(BS ′ A) = kTr(ABS ′ ) = kTr(S2′ ABS2′ ) = Tr(kAB) ∈ Z. Lastly, we check S 3 = I: S 3 = k 3 B 2 S ′ B 2 S ′ B 2 S ′ = k 3 BS ′ ABS ′ ABS ′ A = A−1 S2′ (kAB)3 S2′ A = I.  The following well-known result helps to narrow down the possibilities for S = ρ(s1 s2 ), we include a proof for completeness. Lemma 2.12. Let X ∈ End(V ) with V a d-dimensional vector space. Suppose the minimal polynomial of X coincides with its characteristic polynomial. Then any Y with XY = Y X is of the form: d−1 X bn X n Y = n=0

Proof. The hypothesis implies that there exists a v ∈ V such that the set {v, Xv, · · · , X d−1 v} P n n is a basis for V . Therefore Y v = d−1 n=0 bn X v for some bn . Since Y commutes with X , we Pd−1 have Y = n=0 bn X n because they agree on the basis.  Proposition 2.13. Let ρ be any B3 representation with ρ(σ1 ) = A and ρ(σ2 ) = B such that the characteristic and minimal polynomials of B coincide. Then any extension of ρ to LB 3 has d−1 X ρ(s1 s2 ) = S = an B n AB n=0

for some scalars a0 , . . . , ad−1 . Proof. Since SA = BS by (L1) and AB −1 A−1 = B −1 A−1 B by (B1), we compute: (S(AB)−1 )B = SB −1 A−1 B = SAB −1 A−1 = BSB −1 A−1 = B(S(AB)−1 ). So by Lemma 2.12 S(AB)−1 is a polynomial in B of degree at most d − 1 as required. 7



2.4. Symmetric extensions. Representations of LB3 that factor over SLB 3 contain essentially no topological information, see [13]. We show here that it is not common to find extensions that factor through SLB 3 . Lemma 2.14. A matrix S1 ∈ GLd (C) commutes with a full d × d Jordan block matrix J and S12 = Id if and only if S1 = ±Id .

Proof. Since S1 and J commute, S1 is a polynomial f (J) in J = Id + N where N is the nilpotent Jordan matrix of nilpotency d − 1. Now the result follows from S12 = Id . 

Proposition 2.15. Let ρ be a d-dimensional LB3 representation with ρ(σ1 ) = A and ρ(σ2 ) = B, whose minimal and characteristic polynomials agree, and ρ(s1 ) = S1 and ρ(s2 ) = S2 . Suppose (AB)3 = cId , then ρ factors through SLB 3 if and only if [S2 , B 2 ] = [S1 , A2 ] = 0.

Proof. From Proposition 2.13, S = a0 AB + a1 BAB + · · · + ad−1 B d−1 AB. By Lemma 2.10 (b), S2 commutes with a0 + · · · + ad−1 B d−1 . Now B 2 AB is a multiple of A−1 B −1 when (AB)3 = cId . So we can rewrite S(A−1 B −1 )−1 up to scalar as (a0 + · · · + ad−1 B d−1 )B −2 . Now S2 commutes with the last expression if and only if it commutes with B −2 . Hence if we replace A and B by A−1 and B −1 in Lemma 2.10, we see this is exactly (L2′ ) as required. The condition for A is similar.  Theorem 2.16. Let ρ be a d-dimensional B3 representation with ρ(σ1 ) = A and ρ(σ2 ) = B, whose minimal and characteristic polynomials agree and that no eigenvalue of A is the negation of another. Suppose a standard extension is possible. Then there are only finitely many extensions of ρ that factor through SLB 3 .

Proof. By Proposition 2.15, S1 commutes with A2 . Since no eigenvalue of A is a negation of another, A2 must have a characteristic polynomial equal to its minimal polynomial. In other words, the Jordan blocks have distinct eigenvalues. Restricting S1 to a any A2 eigenspace we apply Lemma 2.14 to see that only finitely many S1 can commute with A2 . Similarly for S2 and B.  We will see in the low-dimensional cases that the assumption on the eigenvalues cannot be removed and that finitely many cannot be strengthened to none. 3. Two-dimensional representations The main goal of this section will be to prove the following: Theorem 3.1. Let (ρ, V ) be any two-dimensional irreducible representation of LB3 with ρ(σ1 ) = A 6= ρ(σ2 ) = B, ρ(s1 ) = S1 and ρ(s2 ) = S2 . Then ρ(s1 s2 ) = −(Tr(AB))−1 AB. Moreover, every representation of B3 with ρ(σ1 ) = A 6= ρ(σ2 ) = B can be extended in this way. Let Gr(V, 1) be the set of one-dimensional linear subspaces of V . Choices for (S1 , S2 ) are in one to one correspondence with Gr(V, 1) \ {Vω , Vω2 }. Remark 3.2. We only consider the A 6= B case because A = B was considered in the previous section. The reducible representations must satisfy S1 = S2 (and therefore A = B) and is a straightforward calculation we have omitted. If A 6= B then we must have S1 6= S2 , whence S3 must act by its two-dimensional irreducible representation so that any extension to LB3 must also be irreducible. We also have Tr(S1 ) = Tr(S2 ) = 0, Tr(S) = −1 and Det(S1 ) = Det(S2 ) = −1. 8

Lemma 3.3. If γ((1 2)) = S1 , γ((2 3)) = S2 defines an irreducible two-dimensional S3 representation, then the matrix equation XS1 = S2 X has a two-dimensional solution space spanned by S1 S2 and S2 S1 S2 . Proof. It is easy to see that S1 S2 and S2 S1 S2 are solutions, which are linearly independent because S1 6= S2 . Now viewing X → XS1 − S2 X as a linear transformation on M2 (C) it suffices to show the image is at least two dimensional. Evaluating at X = 1 and X = S1 gives S1 − S2 and 1 − S2 S1 in the image. They are independent because one has zero trace and the other does not.  Lemma 3.4. If A 6= B satisfy the braid relation (B1) then either     λ1 λ1 λ1 −λ2 A= ,B = , −λ1 /λ2 = ω ±1 , 0 λ2 0 λ2     λ1 λ1 λ2 0 A= ,B = , 0 λ2 −λ2 λ1

or

λ21 − λ1 λ2 + λ22 6= 0.

Moreover, (Tr(AB))2 = Det(AB). Proof. If reducible, then we are in the first case. Otherwise, we are in the second. The last statment is a calculation.  Proof of Theorem 3.1. First we show S = −(Tr(AB))−1 AB. Lemma 3.3 gives AB = c0 S1 S2 + c1 S2 S1 S2 . Taking the trace gives c0 = −Tr(AB) and taking determinant gives Det(AB) = (−Tr(AB) + c1 )(−Tr(AB) − c1 ). Hence by Lemma 3.4, c1 = 0 and S1 S2 = −(Tr(AB))−1 AB. Now we show every B3 representation with A 6= B admits a standard extension. By Lemma 3.4, the characteristic equation for AB is (AB)2 − Tr(AB)AB + (Tr(AB))2 = 0. Multiplying both sides by AB + Tr(AB), we see (−(Tr(AB))−1 AB)3 = 1. For the last claim let Cv be a one-dimensional subspace spanned by v ∈ V , then v = vω + vω2 for unique vω ∈ Vω \ {0} and vω2 ∈ Vω2 \ {0} we define S1 by S1 (vω ) = vω2 . Corollary 3.5. Suppose V is an irreducible two-dimensional representation of LB3 . Then V factors through SLB 3 if and only if Tr(B) = 0. Proof. Suppose Tr(B) = 0. Then from the characteristic equation of B, B 2 is a scalar multiple of I, and hence AB is a scalar multiple of B 2 AB which is a scalar multiple of A−1 B −1 . Conversely, if the opposite relations are satisfied, we get a standard extension of A−1 and B −1 and A and B simultaneously. So AB is a scalar multiple of A−1 B −1 . Therefore B 2 is a scalar multiple of I and from the characteristic equation Tr(B) = 0 since B is not a scalar.  9

4. Three dimensional representations In this section, we show that most irreducible three dimensional representations come from standard extensions in Theorem 4.5. First we state a slightly stronger version of Theorem 2.4 in three dimensions. Lemma 4.1. Let ρ be a three dimensional representation of B3 such that ρ(σ1 ) = A 6= B = ρ(σ2 ). Then there exists a (standard) extension to LB 3 with ρ(s1 s2 ) = S = kAB if and only if Tr(AB) = Tr((AB)2 ) = 0. In this case we have k 3 = Det(AB)−1 . Proof. Suppose there exists a standard extension i.e. with S = kAB. Since A 6= B we have S 6= I. Restricting to S3 and considering characters we see that Tr(S ±1 ) = 0. Therefore Tr(AB) = k −1 Tr(S) = 0 and since (AB)2 is a multiple of S −1 we have Tr((AB)2 ) = 0 as well. Conversely, suppose Tr(AB) = Tr((AB)2 ) = 0. The characteristic polynomial of a 3 × 3 matrix C is −x3 + Tr(C)x2 + [Tr(C)2 − Tr(C 2 )]x + Det(C),

so for C = AB the Cayley-Hamilton Theorem implies (AB)3 = Det(AB)I. Thus for any choice of k 3 = Det(AB)−1 we have (kAB)3 = I so that Theorem 2.4 implies the result.  Proposition 4.2. Let ρ be a three dimensional representation of LB3 . Suppose the minimal and characteristic polynomials of ρ(σ1 ) = A (and ρ(σ2 ) = B) coincide, Tr(AB) = Tr(B 2 AB) = 0, Tr(B 4 AB) 6= 0 and A 6= B. Then S := ρ(s1 s2 ) = kAB.

Proof. Observing that Tr(B 2 AB) = Tr((AB)2 ), we apply Lemma 4.1 to see that there exists a k such that (kAB)3 = I. Thus (BAB)2 is a scalar multiple of I and so Tr(BAB) 6= 0 as ρ is three dimensional. By Proposition 2.13 we have S = a0 AB + a1 BAB + a2 B 2 AB. Since A 6= B and SA = BS, then S 6= I. Therefore, Tr(S) = 0 and taking the trace of the above expression gives a1 = 0. Multiply by B 2 we obtain B 2 S = a0 B 2 AB + a2 B 4 AB.

Noting (by Proposition 2.11) that Tr(B 2 S) = Tr(AB) = 0 = Tr(B 2 AB), we get a2 Tr(B 4 AB) = 0. So a2 = 0, i.e. S = a0 AB.  The following is a summary of the results in [22] on three dimensional B3 representations.

Lemma 4.3 ([22]). Let λ1 , λ2 , λ3 ∈ C× and define:     λ3 0 0 λ1 λ1 λ3 λ−1 2 + λ2 λ2      λ2 0 λ2 λ2  A= .  and B = −λ2 0 λ2 −λ1 λ3 λ−1 0 0 λ3 2 − λ2 λ1

Then: (a) Tr(AB) = Tr(B 2 AB) = Tr((AB)2 ) = 0 whereas Tr(B 4 AB) = λ1 λ2 λ3 (λ1 + λ2 )(λ1 + λ3 )(λ2 + λ3 ) and (AB)3 = (λ1 λ2 λ3 )2 I. (b) ρ(σ1 ) = A and ρ(σ2 ) = B defines a representation of B3 , which is irreducible provided λ2i + λj λk 6= 0 for {i, j, k} = {1, 2, 3}. In this case, the minimal and characteristic polynomials of A and B coincide. 10

(c) Let (ρ, V ) be any irreducible three dimensional representation of B3 . Then there exists a basis of V with respect to which ρ(σ1 ) = A and ρ(σ2 ) = B. (d) Two three-dimensional irreducible representations ρ, ρ′ of B3 are isomorphic if and only if Spec(ρ(σ1 )) = {λ1 , λ2 , λ3 } = Spec(ρ′ (σ1 )).

We may now describe all extensions of three dimensional Tuba-Wenzl representations of B3 to LB3 . In particular we have a full description of extensions of irreducible B3 representations, up to equivalence.

Proposition 4.4. Let (ρ, V ) be a three dimensional representation of B3 such that ρ(σ1 ) = A and ρ(σ2 ) = B as in Lemma 4.3, with Spec(A) = {λ1 , λ2 , λ3 }. Then ρ extends to LB3 , and either   ±1 0 0 (a) ρ(s1 s2 ) = S = γ −2 AB where γ 3 = λ1 λ2 λ3 , ρ(s1 ) = MS1 M −1 where S1 =  0 0 α 0 α1 0 and M ∈ GL(V ) is any matrix such that γ −2 M −1 ABM = diag(1, ω, ω 2), or (b) ρ(s1 s2 ) = S 6= kAB, (λ1 + λ2)(λ1 + λ3 )(λ2 + λ3 ) = 0, and ρ factors over SLB 3 . After permuting  the eigenvalues ifnecessary, we have:   1 z−1 z 0 0 z      and ρ(s1 ) = ± 0  where z ∈ C× is a z z 0 z z ρ(s1 s2 ) =      − z12 free parameter.

1−z 3 z2

0

−z

1−z 2 z

−z

Proof. Combining Lemma 4.3 with Proposition 4.2 we get either S = kAB or (λ1 + λ2 )(λ1 + λ3 )(λ2 + λ3 ) = 0. In the first case the constant is as in Lemma 4.1 and the parametrization of S1 is trivial. Hence we have (a). Assume (λ1 + λ2 )(λ1 + λ3 )(λ2 + λ3 ) = 0 and let S = a0 AB + a2 B 2 AB with a2 6= 0 (see proof of Proposition 4.2 for why a1 = 0). By Lemma 2.10(b) S2 = ρ(s2 ) commutes with S(BA)−1 = a0 + a1 B 2 so S2 commutes with B 2 . Similarly, S1 := ρ(s1 ) commutes with A2 , so by Proposition 2.15 we see that ρ factors over SLB 3 . By Lemma 4.3(d), we may relabel the λi so that λ3 = −λ2 . Set z = λ1 λ2 (a0 + λ22 a2 ). We obtain the required form of S by solving S 3 = I and Tr(ABS) = Tr(AB) and ABS1 = S2 AB.  One consequence of this result is that, generically, three dimensional B3 representations equivalent to the form of Lemma 4.4 do not have non-standard extensions. This is also true in slightly greater generality, as the following illustrates (cf. Corollary 3.5 and Theorem 2.16). Theorem 4.5. Let (ρ, V ) be an irreducible three dimensional representation of LB3 with ρ(σ1 ) = A, ρ(σ2 ) = B with Spec(A) = Spec(B) = {λ1 , λ2 , λ3 } and ρ(s1 s2 ) = S. Then: (a) Any other LB3 representation ψ with ψ(σ1 ) = A, ψ(σ2 ) = B and ψ(s1 s2 ) = S is also irreducible. (b) Suppose the minimal and characteristic polynomials of A (and B) coincide, (AB)3 = cI and (λ1 + λ2 )(λ1 + λ3 )(λ2 + λ3 ) 6= 0. Then ρ(s1 s2 ) = kAB. Proof. For (a), suppose some (ψ, U) satisfying the hypotheses is reducible. Clearly ψ(s1 s2 ) = S = ρ(s1 s2 ) 6= I since A = SA = BS = B implies dim(V ) is even by Theorems 2.7 and 2.8. 11

Thus S has 3 distinct eigenvalues 1, ω ±1, and the only ψ(s1 ), ψ(s2 ) invariant subspaces of U are the 1 eigenspace U1 of S and its S-invariant complement Uω ⊕ Uω−1 . So one of these two spaces is invariant under A, B and S. But then the same is true for (ρ, V ) which contradicts irreducibility as they will also be ρ(s1 ), ρ(s2 ) invariant. For (b), if B3 acts irreducibly on V , then Proposition 4.4 gives the result. So assume that B3 acts reducibly on V and S = ρ(s1 s2 ) 6= kAB for any k. The hypotheses and Proposition 2.13 imply that any A, B invariant subspace W ⊂ V is also S-invariant. By Lemma 4.1 we have S = a0 AB + a2 B 2 AB, from which it follows from Lemma 2.10(b) that S2 = ρ(s2 ) commutes with B 2 . Thus if B 2 has distinct eigenvalues S2 is a polynomial in B 2 by Lemma 2.12 and thus W is S2 invariant, a contradiction. After permuting labels we may assume λ21 = λ22 . We eliminate the possibility that λ1 = λ2 by considering the sizes of the Jordan blocks when the minimal polynomial of B must be of degree 3. Thus (λ1 + λ2 ) = 0, a contradiction.  We illustrate our results with some applications. Example 4.6. The Lawrence-Krammer-Bigelow representation for B3 (see for example [15]) is defined by     2 1−q 0 1 tq 0 tq(q − 1)     2 2 ,  , ρ(σ2 ) = B =  0 tq tq (q − 1) 0 1 − q q ρ(σ1 ) = A =      q 0 0 0 1 0

for q, t ∈ C× . We see that Tr(AB) = Tr(B 2 AB) = 0 and a standard extension is possible by Lemma 4.1. The minimal polynomial of A and B both have degree 3 and the eigenvalues are not negations of each other when tq 2 6= −1, tq 6= 1 and q 6= 1. In this case, the irreducible extensions must be standard by Theorem 4.5. See also [3], where it is shown that the Lawrence-Krammer-Bigelow representation of Bn with n ≥ 4 does not extend except for degenerate cases.   0 t 0 Example 4.7. For any 1 6= t ∈ C, the B3 representation given by ρ(σ1 ) = A = 1 0 0 0 0 1   1 0 0 and ρ(σ2 ) = B = 0 0 t  is irreducible (see [7]). Thus there is a basis with respect to 0 1 0 √ which A and B are in the Tuba-Wenzl form. The eigenvalues of A are {1, ± t}, so that a 1-parameter family of non-standard   extensions exist, byProposition  4.4(b). It is easy to 0 1 0 1 0 0    verify that ρ(s1 ) = S1 = 1 0 0 and ρ(s2 ) = S2 = 0 0 1 defines an extension. 0 0 1 0 1 0   2 0 0 t  Since AB = 1 0 0  is is clear that ρ(s1 s2 ) 6= kAB for this example. One may verify 0 1 0 directly that this representation factors over SLB3 . In fact, the analogous extension works for all n and factors over SLB n . 12

5. Dimensions 4 and 5 The following is a summary of the results in [22] on four dimensional B3 representations. Lemma 5.1 ([22]). Let λ1 , λ2 , λ3 , λ4 ∈ C× and choose γ 2 such that γ 4  λ1 λ4 λ21 λ24 λ1 λ4 λ21 λ24 λ1 (1 + γ 2 + γ 4 )λ2 (1 + γ 2 + γ 4 )λ3    λ1 λ4  λ2 (1 + 2 )λ3 A=0 γ   0 0 λ3  0 0 0

= λ1 λ2 λ3 λ4 . Define  λ4     λ4     λ4   λ4

and 

λ4

0

0

0



   −λ3  λ 0 0 3      λ2 λ3  λ2 λ3 2 B=  −( + 1)λ λ 0 2 2  γ2  2 γ     2 2  λ1 λ3 λ3 λ3 λ3 λ2 λ2 λ2 λ3  λ2 λ3 λ2 λ3 2 3 2 3 2 3 ( + + )λ −( + + 1)λ λ − 1 1 1 γ6 γ6 γ4 γ2 γ4 γ2 (1) Setting ρ(σ1 ) = A and ρ(σ2 ) = B defines a representation of B3 , and every irreducible B3 representation of dimension 4 is equivalent to such a representation. (2) Tr(AB) = −γ 2 and (AB)3 = −γ 6 I4 . Proposition 5.2. All four dimensional Tuba-Wenzl representations admit a standard extension. In particular, all irreducible four dimensional representations of B3 admit a standard extension. Proof. By Lemma 5.1, (−γ −2 AB)3 = I4 and Tr(−γ −2 AB) = 1. Now apply Theorem 2.4.  Proposition 5.3. Any extension of a four dimensional Tuba-Wenzl representation satisfies Tr(S) = 1. Proof. Tr(S) must either be −2 or 1 since S 6= 1. If Tr(S) = −2, then e1 + Se1 + S 2 e1 = 0 and e2 + Se2 + S 2 e2 = 0 means S must have diagonal entries 0, 0, −1, −1 and other entries zero unless they are on the skew diagonal, in which case they would be x, x, −x−1 , −x−1 . We see that such a matrix cannot satisfy SA = BS. Equating the (2, 4) entry gives λ4 = −λ3 . Then equating the (3, 4) entry gives x(x − 1) = λ22 /γ 2 .  Example 5.4. If V and W are both two-dimensional irreducible representations of LB 3 then V ⊗ W is a four dimensional representation with S = kAB. The following is a summary of the classification of simple B3 representations of dimension 5 found in [22]. Lemma 5.5 ([22]). Let λ1 , λ2 , λ3 , λ4 , λ5 ∈ C× with γ a fixed fifth root of λ1 λ2 λ3 λ4 λ5 . Define 13



 λ1    0   A=  0    0  0

 γ2 γ3 λ1 λ5 λ2 λ4 γ3 γ2 γ3 (1 + )(λ2 + ) (1 + 2 )(λ3 + γ + ) (1 + 2 )(λ3 + ) λ2 λ4 λ3 λ4 γ λ3 γ λ2 λ4 λ1 λ5    3 3  2 γ γ  γ λ3 + γ + λ2 λ3 + γ + λ3 λ1 λ5 λ1 λ5    3  3 γ  γ 0 λ3 λ3 + λ1 λ5 λ1 λ5    0 0 λ4 λ4   0 0 0 λ5

and B by Bi,j = (−1)i−j A6−i,6−j . (1) Defining ρ(σ1 ) = A and ρ(σ2 ) = B defines a representation of B3 , and every irreducible B3 representation of dimension 5 is equivalent to such a representation. (2) (γ −2 AB)3 = I5 and Tr(γ −2 AB) = −1. From this lemma and Theorem 2.4 we immediately have:

Proposition 5.6. All five-dimensional Tuba-Wenzl representations admit a standard extension. In particular, all irreducible five-dimensional representations of B3 admit a standard extension. Proposition 5.7. Any extension of a five-dimensional Tuba-Wenzl representation satisfies Tr(S) = −1.

Proof. Assume otherwise, Tr(S) = 2. In this case the non real eigenspace of S is a unique two-dimensional subspace. So the span of e1 , Se1 , S 2 e1 and e2 , Se2 , S 2 e2 must have a twodimensional intersection. Assuming the skew triangular form of S, this determines the 3rd row and column completely (all but (3,3) entry zero, which is 1). Applying SA = BS to this row and column gives two of the skew diagonal entries next to the (3,3) entry are both -1. Now S 3 e1 = e1 and S 3 e2 = e2 gives the matrix for S has only non-zero entries on the skew diagonal x, −1, 1, −1, x−1 and the last row and column: x, v, 0, v, 1, x−1 , −v −1 , 0, −v −1, 1. This now should be checked to fail the relation SA = BS.  In dimension d = 4 and 5, every irreducible B3 representation can be extended to a standard LB3 representation. Now we show that most extensions are standard. That is, whenever an extension exists, S = kAB for some k ∈ C× unless the eigenvalues (λ1 , . . . , λd ) of A and B are zeros of a set of polynomials. Proposition 5.8. Let ρ be an extension to LB3 of an irreducible B3 representation of dimension 4 or 5 in the form of Lemma 5.1 or 5.5, with ρ(σ1 ) = A, ρ(σ2 ) = B and ρ(s1 s2 ) = S. Then AB, S and BSA are all skew lower triangular. Furthermore, S 2 and (BSA)2 are skew upper triangular matrices.

Proof. A direct computation shows that AB is skew lower triangular. Since the characteristic and minimal polynomials of A coincide, Proposition 2.13 implies S is the product of a lower triangular matrix by a skew lower triangular matrix, which is evidently skew lower triangular. Therefore, S 2 = S −1 is skew upper and the same holds for BSA = B 2 A by Proposition 2.11.  14

Theorem 5.9. Let ρ be an irreducible d-dimensional B3 -representation with d = 4 or 5 as in Lemmas 5.1 and 5.5 with ρ(σ1 ) = A and ρ(σ2 ) = B. For each d = 4 and 5, there is a variety V (Jd ) where Jd ⊂ K[λ1 , . . . , λd , γ], such that if (λ1 , . . . , λd ) ∈ Cd \ V (Jd ), then S = kAB is the only solution for the loop braid relations, where k = −γ −2 for d = 4 and k5 = γ −2 for d = 5. P i Proof. By Proposition 2.13, S = d−1 i=0 bi B AB. We will prove that, except when (λ1 , . . . , λd , γ) is a solution to a certain system of polynomial equations, all coefficients bi vanishes except b0 . Let Pij be the (i, j)-entry of S 2 and Qij be the (i, j)-entry of (BSA)2 . Then Pij and Qij are second degree homogeneous polynomials of b0 , . . . , bd−1 . Proposition 5.8 gives Pij = 0 and Qij = 0, if i < j

(∗)

Note that for i < j, the term b20 does not appear in Pij and Qij for (AB)2 and (BABA)2 are skew upper triangular. Therefore, for i < j, System (∗) can be viewed as a homogeneous linear system of unknowns bm bn . It has more equations than unknowns when d ≥ 2. In fact, it has d2 − d equations and Nd unknowns bm bn (m + n > 0), where Nd = (d + 2)(d − 1)/2 and Nd ≤ d2 − d for d ≥ 2. In general, it only has zero solutions depending on whether the coefficient matrix has rank Nd . Let Md be the coefficient matrix of System (∗) and Jd be the set of the determinants of all Nd × Nd submatrices of Md . If (λ1 , . . . , λd , γ) ∈ / V (Jd ), then System (∗) only has zero solutions, that is bm bn = 0 for m + n > 0. Therefore, for generic values of (λ1 , . . . , λd , γ) (i.e. away from V (Jd ) we must have S = b0 AB. 

6. Beyond dimension 5 We show that six-dimensional irreducible representations of B3 do not always extend. Proposition 6.1. Let ω = e2πi/3 be a primitive 3rd root of unity and define: 

1

−ω + 1

 2  ω −1 ω2   2  ω −1 ω2 − 1  A=  0 ω−1   −ω 2 + 1 −ω 2 + 1 

−ω 2 + 1

ω−1

ω2 − 1

0

−ω 2 + 1

0

0

ω2 − 1

ω2 − 1

−1

ω2 ω2 − 1 0

−ω 2 + 1 −ω 2 + 1 −ω 2 + 1 15

ω−1



    −ω 2 + 1 −ω 2 + 1 0    2 −ω −ω + 1 −ω + 1   ω2 − 1 −1 0  

and  −ω + 1 −ω 2 + 1 −ω + 1 −ω 2 + 1 −ω + 1   2 2 2  ω − 1 ω 0 ω − 1 0 0     2 2 2 2  ω − 1 ω 2 − 1 ω ω − 1 ω − 1 0   B= . 2 2  0 −ω + 1 −ω + 1 −ω −ω + 1 −ω + 1     2  ω 2 − 1 ω 2 − 1 0 ω − 1 −1 0   

1

ω2 − 1

ω2 − 1

ω2 − 1

ω2 − 1

ω2 − 1

−1

Then ρ(σ1 ) = A and ρ(σ2 ) = B define an irreducible representation of B3 which cannot be extended to an LB3 representation. Proof. These two matrices follows the construction in [17], where the representation is shown to be indecomposable. We verify (using Magma [4]) that the dimension of the algebra generated by A, B is 36, we see that the representation is irreducible. Alternatively, one may use [22, Remark 2.11(4)] to verify irreducibility: in this case Det(A)6 = 1 so that if there were a non-trivial r < 6-dimensional subrepresentation W ⊂ C6 , the eigenvalues of A on W must satisfy (µ1 · · · µr )36 = 1. Computing the characteristic polynomial one finds that √ the πi/3 eigenvalues of A are e and the 5 roots of an irreducible 5 degree polynomial in Q( 3i)[x]. In particular, the minimal polynomial of B (and A) coincides with the characteristic polynomial. In any case, any subrepresentation either has dimension 1 or a 1 direct complement. One then verifies that A and B do not have a common eigenvector of eigenvalue eπi/3 . P5 Now if an extension of ρ to LB3 exists, then, by Proposition 2.13, S = i=0 bi B i AB. The requirement S 3 = I and a calculation imply that S = qAB or S = qB 2 AB where q is any third root of unity. But for such S, tr(S) ∈ / R. Therefore, this irreducible B3 representation cannot be extended to a LB3 representation.  In [22, Subsection 1.5] there is a general construction of (d + 1) × (d + 1) representations in ordered triangular form. We recall the details for the reader’s convenience. For 0 ≤ i ≤ d define i = d−i and let c ∈ C× . For 0 ≤i ≤ and  matrices   A   d, let λi satisfy λi λd−i = c. Define   P i i i i k λi . The identity: dk=0 (−1)k λi and Bij = (−1)i+j B by Aij = = is k j j j j useful to verify the relation ABA=BAB so that these matrices define a B3 representation. d i and (AB)3 = (−1)d c3 I. Define S = (−1) AB, that is: One computes: (AB)ij = c(−1)j c j     i i j d+j = (−1) . Sij = (−1) j j Clearly Tr(S) ∈ R (in fact Tr(S) ∈ {0, ±1} and S 3 = I) and this representation admits a standard extension. Using [22, Remark 2.11(4)] (due to Deligne) it is possible to show that, for sufficiently generic eigenvalues, these B3 representations are irreducible. Our investigations of low-dimensional LB3 representations suggest the following: Conjecture 1. Suppose that ρ is an irreducible d-dimensional matrix representation of B3 such that ρ(σ1 ) = A and ρ(σ2 ) = B are in ordered triangular form ([22]), that is, A is 16

upper triangular and B is lower triangular with Bi,i = Ad−i+1,d−i+1 . Then ρ has a standard extension to LB3 . To prove this conjecture it is enough to show that for some root k of cx3 − 1 where (AB)3 = cI we have Tr(kAB) ∈ Z. Fixing some root c1/3 , we have Spec(c1/3 AB) ⊂ {1, ω ±1} where ω = e2πi/3 . If these appear with multiplicities µ1 , µ± then there is a choice of k such that Tr(kAB) ∈ Z if and only if two of these eigenvalues coincide. References [1] J. C. Baez, D. K. Wise, A. S. Crans, Exotic statistics for strings in 4D BF theory. Adv. Theor. Math. Phys. 11 (2007), no. 5, 707–749. [2] V. G. Bardakov, The structure of a group of conjugating automorphisms. Algebra Logika 42 (2003), no. 5, 515–541, 636; translation in Algebra Logic 42 (2003), no. 5, 287–303. [3] V. G. Bardakov, Extending representations of braid groups to the automorphism groups of free groups. J. Knot Theory Ramifications 14 (2005), no. 8, 1087–1098. [4] W. Bosma, J. Cannon, and C. Playoust, The Magma algebra system I: The user language. J. Symbolic Comput. 24 (1997), 235–265. [5] D. M. Dahm, A generalization of braid theory, PhD thesis, Princeton University, 1962. [6] R. Fenn, R. Rim´anyi, C. Rourke, Some remarks on the braid-permutation group. Topics in knot theory (Erzurum, 1992), 57–68, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 399, Kluwer Acad. Publ., Dordrecht, 1993. [7] E. Formanek, W. Lee, I. Sysoeva, M. Vazirani, The irreducible complex representations of the braid group on n strings of degree ≤ n. J. Knot Theory Ramifications 2 (2003), no. 3, 317–333. [8] M. Freedman; A. Kitaev; M. Larsen; Z. Wang, Topological quantum computation. Bull. Amer. Math. Soc. (N.S.) 40 (2003) no. 1, 31–38. [9] M. H. Freedman; M. J. Larsen; Z. Wang, A modular functor which is universal for quantum computation. Comm. Math. Phys. 227 (2002) no. 3, 605–622. [10] F. Gantmacher, The Theory of Matrices, Chelsea, New York, 1959. [11] D. L. Goldsmith, The theory of motion groups. Michigan Math. J. 28 (1981), no. 1, 3–17. [12] Z. Kadar, P. Martin, E. Rowell, Z. Wang, Local representations of the loop braid group, preprint arXiv:1411.3768 (2015). [13] T. Kanenobu, Forbidden moves unknot a virtual knot, J. Knot Theory Ramifications 10 (2001) no. 1, 89–96. [14] L. H. Kauffman, Virtual Knot Theory. European J. Comb. 20 (1999), 663–691. [15] D. Krammer Braid groups are linear. Annals of Mathematics 155 (2002), 131–156. [16] M. J. Larsen, E. C. Rowell, Z. Wang, The N -eigenvalue problem and two applications. Int. Math. Res. Not. 64 (2005), 3987–4018 [17] L. LeBruyn, Rationality and dense families of B(3) representations, preprint arXiv:1003.1610 (2010). [18] X.S. Lin, The motion group of the unlink and its representations. Topology and physics : Proceedings of the Nankai International Conference in Memory of Xiao-Song Lin, Tianjin, China, 27-31 July 2007. Editors, Kevin Lin, Zhenghan Wang, Weiping Zhang. World Scientific, 2008. [19] J. McCool, On basis-conjugating automorphisms of free groups. Canad. J. Math. 38 (1986), no. 6, 1525–1529. [20] L. L. Pennisi, Coefficients of the Characteristic polynomial. Mathematics Magazine 60 (1987), no. 1, 31–33. [21] E. C. Rowell, Braid representations from quantum groups of exceptional Lie type, Rev. Un. Mat. Argentina 51 (2010) no. 1, 165–175. [22] I. Tuba, H. Wenzl, Representations of the braid group B3 and of SL(2, Z). Pacific Journal of Mathematics 197 (2001), 491–510. [23] V. V. Vershinin, On homology of virtual braids and Burau representation. J. Knot Theory Ramifications 10 (2001), no. 5, 795–812. Knots in Hellas-98, Vol. 3 (Delphi). [24] K. Walker, Z. Wang, (3 + 1)-TQFTs and topological insulators. Front. Phys. 7 (2) (2012), 150–159. 17

[25] C. Wang, M. Levin, Braiding Statistics of Loop Excitations in Three Dimensions Phys. Rev. Lett. 113, 080403. [26] J. C. Wang, X.G. Wen, Non-Abelian string and particle braiding in topological order: Modular SL(3, Z) representation and (3+1)-dimensional twisted gauge theory Phys. Rev. B 91, 035134 [27] L. Zadeh, C. Desoer, Linear System: The State Approach, McGraw-Hill, New York, 1963. E-mail address: [email protected] Pacific Northwest National Laboratory, 902 Battelle Boulevard, Richland, WA U.S.A E-mail address: [email protected] Department of Mathematics and Statistics, University of Toledo, Toledo, OH U.S.A. E-mail address: [email protected] ´tica, Astronom´ıa y F´ısica, Universidad Nacional de Co ´ rdoba, CIEM Facultad de Matema ´ rdoba, Argentina. – CONICET, (5000) Ciudad Universitaria, Co E-mail address: [email protected] Department of Mathematics, Texas A&M University, College Station, TX U.S.A. E-mail address: [email protected] Department of Mathematics, Texas A&M University, College Station, TX U.S.A. E-mail address: [email protected] ¨t Mu ¨nster, Germany Mathematisches Institut der Universita

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